\documentclass[lang=cn,11pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{ctex}
\title{数值分析作业1}
\author{庞竣元 \qquad 3200105017}
\date{}

\begin{document}

\maketitle

\noindent\uppercase\expandafter{\romannumeral1}.\par
\noindent(1)\ $\frac{1}{2^{n-1}}$\par
\noindent(2)
第一次迭代时取值为区间中点，故距离零点的最长距离为$1$，准确地说，小于$1$\par
\noindent 考虑第二次迭代所选区间内必定包含零点，故从第二步迭代开始，距离最近零点的距离都不会超过1，故我们记这个距离为$r(f)$,有
$$
\sup\{r(f)\}\leq 1 
$$
下面我们证
$$
\sup\{r(f)\}= 1 
$$
任取$d<1$，记停机准则为$\varepsilon >0$，为方便讨论我们不妨假设所在区间为$[0,2]$，
记
\begin{equation*}
    f_d(x)=\left\{\begin{aligned}
             -&\frac12 \varepsilon + \frac{1}{1-d} \varepsilon x  \qquad x \leq 1-d  \\  
             &\frac12 \varepsilon \qquad otherwise   
             \end{aligned}\right 
\end{equation*}


可知$f_d$的零点为$\frac{1}{2}(1-d)$，最终迭代结果为$x_0=1$，故$r(f)=d+\frac{1}{2}(1-d)>d$，证毕。

\noindent\uppercase\expandafter{\romannumeral3}.\par
\noindent 记$x_n$为$f$的第$n$步迭代结果
$$\frac{b_0-a_0}{2^n x_0}<\frac{b_0-a_0}{2^n a_0}<\epsilon$$
有,
\begin{equation*}
    \begin{aligned}
        &\frac{b_0-a_0}{a_0\varepsilon}&<2^n\\
        log(b_0-a_0)-log&(\varepsilon)-log(a_0)&<n\\
        log(b_0-a_0)-log&(\varepsilon)-log(a_0)-1&\leq n
    \end{aligned}
\end{equation*}
\noindent\uppercase\expandafter{\romannumeral3}.\par
\begin{equation*}
    \begin{aligned}
    &p(x) = 4x^3 - 2x^2 + 3 \\
    &p'(x)= 12x^2 - 4x \\
    &x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \\
\end{aligned}
\end{equation*}

得$x^* \approx -0.7688$。

\noindent\uppercase\expandafter{\romannumeral4}.\par

\begin{equation*}
    \begin{aligned}
    x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_0)} \\
    e_{n+1} = f(x_{n+1}) &= f(x_n) + f'(x_n) (x_{n+1}-x_n) + O(x_{n+1}-x_n) \\
        &= f(x_n) - \frac{f'(x_n)\cdot f(x_n)}{f'(x_0)} + O(x_{n+1}-x_n) \\
        &= (1-\frac{f'(x_n)}{f'(x_0)})\cdot e_n \\
\end{aligned}
\end{equation*}

故$s=1,C=1-\frac{f'(x_n)}{f'(x_0)}$。

\noindent\uppercase\expandafter{\romannumeral5}.\par

\begin{equation*}
    \begin{aligned}
    x_{n+1} &= \arctan x_n \\
    g(x) &= \arctan(x)\\
    \end{aligned}
\end{equation*}
考虑到$g : [-\frac{\pi}{2},\frac{\pi}{2}] \rightarrow [-\frac{\pi}{2},\frac{\pi}{2}]$紧致
且迭代序列$\{x_n\}$有下界$0$，故在不动点$x=0$处收敛。\\$x_0=0$，则 $x_n\equiv 0\$。


\noindent\uppercase\expandafter{\romannumeral6}.\par
\noindent 易知$x_{n+1} = \frac{1}{p+x_{n}} $, 初始条件 $ x_1 = \frac{1}{p}$。\\
考虑$g(x)=\frac{1}{p+x},\ [0,1] \rightarrow [0,1]$紧致。
  \begin{equation*}
    \lambda = \max_{x\in[0,1]} |g'(x)| = \max_{x\in[0,1]} -\frac{1}{(x+p)^2} = \frac{1}{p^2} < 1
  \end{equation*}
故$g$紧致，考虑
  \begin{equation*}
    x=g(x)=\frac{1}{p+x}
  \end{equation*}
两根为$\frac{-p\pm\sqrt{p^2+4}}{2}$, 我们考虑正不动点$x_0=\frac{-p+\sqrt{p^2+4}}{2}$，迭代会收敛到该点。
\newpage

\noindent\uppercase\expandafter{\romannumeral7}.\par
 \begin{equation*}
    \begin{aligned}
        a_0 < &0 < b_0 \\
    |x_n - x^*| &\leq \frac{b_0-a_0}{2^n} \\
       \varepsilon = \frac{|x_n-x^*|}{|x^*|} &\leq \frac{b_0-a_0}{2^n\cdot|x^*|} \\
    \end{aligned}
\end{equation*}
故  $ n  \geqslant log_2(b_0-a_0) - log_2\varepsilon -log_2|x^*| $ 时,
      相对误差小于 $\varepsilon$ 。\\

\end{document}
